Integrating factors problem

  • Thread starter ssb
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  • #1
ssb
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Homework Statement



[tex]y' + 2ty = t^3[/tex]

Homework Equations



Integrating factors and variation of parameters

The Attempt at a Solution



Ive solved for m
[tex] M = e^{\int 2t\,dx} [/tex]

[tex] M = e^t^2 [/tex] (this is e^t^2, but doesn't look like it in latex)

I multiplied both sides by M

[tex](e^t^2)(y') + (e^t^2)(2ty) = (e^t^2)(t^3)[/tex]

The part I am having trouble with is the next step. I am suppost to find a function so when you take the derivative of that function, it is the same as the left hand side of this problem, or [tex](e^t^2)(y') + (e^t^2)(2ty)[/tex]

Any suggestions? Is there a simple formula you can use that always works in these cases? It seems like I've tried [tex]\frac{d}{dt}(y)(e^t^2)[/tex] but it won't work.
 
Last edited:

Answers and Replies

  • #2
bdeln
7
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Just pick the undifferentiated part from each term on the left hand side, ie,

[itex]\left( e^{t^2} y \right)' = t^3 e^{t^2}[/itex]
 
  • #3
ssb
119
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Just pick the undifferentiated part from each term on the left hand side, ie,

[itex]\left( e^{t^2} y \right)' = t^3 e^{t^2}[/itex]

Thats exactly what I've been doing yet webworks is rejecting my work. Let me provide the rest of my steps just to verify:

You know what... since I am over the deadline webworks gives me the option to show answers... my answers I am suppost to be putting in are all jacked up. Looks like the teacher messed up programming it this week for me. I've been doing it right all along!!! Thanks for the help buddy!
 
  • #4
ssb
119
0
What if I had

[tex]t^3y' - (3/t)yt^3 = t^6 [/tex]


[tex](?)' = t^6[/tex]


would it be [tex]yt^3[/tex] ????
 
  • #5
bdeln
7
0
Well, we can check ... [itex](t^3y)' = 3t^2y + t^3y'[/itex], so other than the minus sign, I think you're good.
 

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